Question

Liquid octane CH3CH26CH3 reacts with gaseous oxygen gas O2 to produce gaseous carbon dioxide CO2 and gaseous water H2O. If 91.0g of water is produced from the reaction of 97.09g of octane and 598.4g of oxygen gas, calculate the percent yield of water. Round your answer to 3 significant figures.

Answer 1

Molar mass of C8H18,
MM = 8*MM(C) + 18*MM(H)
= 8*12.01 + 18*1.008
= 114.224 g/mol


mass(C8H18)= 97.09 g

use:
number of mol of C8H18,
n = mass of C8H18/molar mass of C8H18
=(97.09 g)/(1.142*10^2 g/mol)
= 0.85 mol

Molar mass of O2 = 32 g/mol


mass(O2)= 598.4 g

use:
number of mol of O2,
n = mass of O2/molar mass of O2
=(5.984*10^2 g)/(32 g/mol)
= 18.7 mol
Balanced chemical equation is:
2 C8H18 + 25 O2 ---> 18 H2O + 16 CO2


2 mol of C8H18 reacts with 25 mol of O2
for 0.85 mol of C8H18, 10.62 mol of O2 is required
But we have 18.7 mol of O2

so, C8H18 is limiting reagent
we will use C8H18 in further calculation


Molar mass of H2O,
MM = 2*MM(H) + 1*MM(O)
= 2*1.008 + 1*16.0
= 18.016 g/mol

According to balanced equation
mol of H2O formed = (18/2)* moles of C8H18
= (18/2)*0.85
= 7.65 mol


use:
mass of H2O = number of mol * molar mass
= 7.65*18.02
= 1.378*10^2 g

% yield = actual mass*100/theoretical mass
= 91*100/1.378*10^2
= 66.03%
Answer: 66.0 %